Verified answer

e^{x^2 -y^2} cos(2xy) =1

e^{x^2 -y^2} sin(2xy) = 0

x=sqrt(pi)=y

then

e^{pi^2 -pi^2} =1,

so cos (2pi)=1 and sin (2pi) =0.

therefore (√π , √π) is a point of intersection.

the derivative of the first curve is:

e^{x^2 -y^2} (-sin(2xy))(2xy'+2y) + e^{x^2 -y^2}(2x- 2yy') cos2xy=0

the derivative of the second curve is:

e^{x^2 -y^2} (cos(2xy))(2xy'+2y) + e^{x^2 -y^2}(2x-2yy') sin2xy=0

if x= √π =y

these become:

(2√π -2√πy') =0

2√πy'+2√π =0

so the slope of the first equation is: y'= 2√π/2√π = 1

and the slope of the second equation is: y'= -2√π/2√π =-1

therefore the 2 tangent lines are perpendicular. .

Well, first show that they both go through that point.

You should be able to do this with simple substitution.

Then show that the slope of each at that point are negative reciprocals of each other to show orthogonality.

Take the derivative of each, and evaluate at x=y=√π

(e^((x^2) -(y^2)))(2x-2yy')cos(2xy)

- (e^((x^2) -(y^2)))sin(2xy)(2xy' + 2y) = 0

and

(e^((x^2) -(y^2)))(2x-2yy')sin(2xy)

+ (e^((x^2) -(y^2)))cos(2xy)(2xy' + 2y) = 0

at x=y=√π, the first becomes:

e^0 *(2√π-2√πy')*cos(2π) + e^0 *sin(2π)(2√π+2√πy')=0

or

2√π - 2√πy' = 0 or y' = 1

and the second becomes:

e^0 *(2√π-2√πy')*sin(2π) + e^0 *cos(2π)(2√π+2√πy')=0

or

2√π + 2√πy' = 0 or y' = -1

Since 1 = -[1/(-1)] they are orthogonal at that point.